Sea water has about 0.46 moles NaCl and 0.065 moles of MgCl₂ in every litre. What is the vapour pressure of sea water at 30?C if pure water would have a vapour pressure of 33.2 mm Hg? (Assume there are 1000 grams of water in 1 litre of sea water.)

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The vapor pressure of a solution can be calculated using Raoult's Law, which states that the vapor pressure of a solution is directly proportional to the mole fraction of the solvent. In this case, the solvent is water. First, we need to calculate the total moles of solute (NaCl and MgCl2) and then the moles of water in 1 liter.Given:- Moles of NaCl = 0.46- Moles of MgCl2 = 0.065- 1000 grams of water = 1000 g / 18.015 g/mol (molar mass of water) = 55.51 moles of waterTotal moles of solute = 0.46 + 0.065 = 0.525 molesTotal moles in solution = 55.51 (water) + 0.525 (solute) = 56.035 molesMole fraction of water = moles of water / total moles in solution = 55.51 / 56.035 ≈ 0.9907Vapor pressure of sea water = mole fraction of water * vapor pressure of pure water = 0.9907 * 33.2 mm Hg ≈ 32.9 mm HgHowever, due to the presence of a typographical error in the original question (MgCl₂), which seems to be an incorrect representation of MgCl2, and considering the closest match to the calculated value, the correct choice based on the provided options and the typical approach to such problems would be:E) 32.9 mm Hg.

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